Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 19: Problem 44

A straight wire segment of length \(0.60 \mathrm{m}\) carries a current of $18.0 \mathrm{A}$ and is immersed in a uniform external magnetic ficld of magnitude \(0.20 \mathrm{T}\). (a) What is the magnitude of the maximum possible magnetic force on the wire segment? (b) Explain why the given information enables you to calculate only the maximum possible force.

Short Answer

Expert verified
Answer: The maximum magnetic force on the wire segment is 2.16 N. Only the maximum possible force can be calculated because we do not have the information about the angle between the current and the magnetic field. However, since we know the maximum value of sinθ is 1 (when θ = 90°), we can calculate the maximum possible force considering the angle to be 90°. Any other angle would result in a lesser force, but we do not have enough information to find the exact force.

Step by step solution

01

Identify given variables

We are given the following values: - Length of the wire (L) = \(0.60 m\) - Current in the wire (I) = \(18.0 A\) - Magnitude of the uniform external magnetic field (B) = \(0.20 T\)
02

Calculate the maximum force

We can calculate the magnetic force on the wire segment using the formula: $$ F = ILB \sin\theta $$ To find the maximum value for the force, observe that \(\sin\theta\) can take values between -1 and 1, with the maximum value being \(\sin {90^\circ} =1\). So, to find the maximum force, set \(\theta = {90^\circ}\): $$ F_{max} = ILB\sin{90^\circ} = ILB $$ Now, plug in the given values and compute F_max: $$ F_{max} = (18.0\,\mathrm{A})(0.60\,\mathrm{m})(0.20\,\mathrm{T}) = 2.16\,\mathrm{N} $$ The maximum possible magnetic force on the wire segment is \(2.16\,\mathrm{N}\).
03

Explain why only the maximum possible force can be calculated

The given information does not provide the angle between the current and the magnetic field, which means the actual force could be lower than the maximum possible force, depending on the value of \(\theta\). However, since we know the maximum value of \(\sin\theta\) is 1, we can calculate the maximum possible force when \(\theta=90^\circ\). Any other angle between current and magnetic field would result in a lesser force. And thus, only the maximum possible force can be calculated with the given information.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

An electromagnet is made by inserting a soft iron core into a solenoid. The solenoid has 1800 turns, radius \(2.0 \mathrm{cm},\) and length $15 \mathrm{cm} .\( When \)2.0 \mathrm{A}$ of current flows through the solenoid, the magnetic field inside the iron core has magnitude 0.42 T. What is the relative permeability \(\overline{K_{\mathrm{B}}}\) of the iron core? (See Section 19.10 for the definition of \(\boldsymbol{K}_{\mathbf{B}} .\) )
The intrinsic magnetic dipole moment of the electron has magnitude $9.3 \times 10^{-24} \mathrm{A} \cdot \mathrm{m}^{2} .$ In other words, the electron acts as though it were a tiny current loop with $N I A=9.3 \times 10^{-24} \mathrm{A} \cdot \mathrm{m}^{2} .$ What is the maximum torque on an electron due to its intrinsic dipole moment in a \(1.0-T\) magnetic field?
Two identical bar magnets lic next to one another on a table. Sketch the magnetic ficld lines if the north poles are at the same end.
An electron moves at speed \(8.0 \times 10^{5} \mathrm{m} / \mathrm{s}\) in a plane perpendicular to a cyclotron's magnetic field. The magnitude of the magnetic force on the electron is \(1.0 \times 10^{-13} \mathrm{N}\) What is the magnitude of the magnetic field?
The conversion between atomic mass units and kilograms is $$1 \mathrm{u}=1.66 \times 10^{-27} \mathrm{kg}$$ A sample containing sulfur (atomic mass 32 u), manganese \((55 \mathrm{u}),\) and an unknown element is placed in a mass spectrometer. The ions have the same charge and are accelerated through the same potential difference before entering the magnetic field. The sulfur and manganese lines are separated by \(3.20 \mathrm{cm},\) and the unknown element makes a line between them that is \(1.07 \mathrm{cm}\) from the sulfur line. (a) What is the mass of the unknown clement? (b) Identify the element.
See all solutions

Recommended explanations on Physics Textbooks

Thermodynamics

Read Explanation

Waves Physics

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Force

Read Explanation

Mechanics and Materials

Read Explanation

Wave Optics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free